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(D) Let us simplify the circuit from right to left.$1$. The last section consists of a $1 \ \Omega$ resistor in series with another $1 \ \Omega$ resis

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$2$

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(D) Let us simplify the circuit from right to left.$1$. The last section consists of a $1 \ \Omega$ resistor in series with another $1 \ \Omega$ resistor,which is in parallel with a $2 \ \Omega$ resistor. The series combination is $1 + 1 = 2 \ \Omega$. This $2 \ \Omega$ is in parallel with the $2 \ \Omega$ resistor: $R_{p1} = \frac{2 	imes 2}{2 + 2} = 1 \ \Omega$.$2$. Now,this $1 \ \Omega$ is in series with the next $1 \ \Omega$ resistor,giving $1 + 1 = 2 \ \Omega$. This is in parallel with the next $2 \ \Omega$ resistor: $R_{p2} = \frac{2 	imes 2}{2 + 2} = 1 \ \Omega$.$3$. Again,this $1 \ \Omega$ is in series with the next $1 \ \Omega$ resistor,giving $1 + 1 = 2 \ \Omega$. This is in parallel with the first $2 \ \Omega$ resistor: $R_{p3} = \frac{2 	imes 2}{2 + 2} = 1 \ \Omega$.$4$. Finally,this $1 \ \Omega$ is in series with the first $1 \ \Omega$ resistor connected to terminal $X$. Total resistance $R_{eq} = 1 + 1 = 2 \ \Omega$.

Net resistance of the given circuit between $X$ and $Y$ is ............... $\Omega$.

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